Let's say there is an algorithm that runs on a processor at x MIPS, operating at a clock frequency of y MHz. If I double the operating frequency to 2y, what happens to the MIPS? It seems to me that doubling the operating frequency should double the instructions executed per second, i.e, MIPS = 2x, is this correct?


Only if you double the speed of everything. Which you can’t in practice.

| cite | improve this answer | |
  • $\begingroup$ Since MIPS probably disregard memory or other I/O, I don't follow. $\endgroup$ – Raphael Dec 21 '17 at 9:48
  • $\begingroup$ MIPS don't have a choice. If you read from or write to memory, it takes as long as it takes. $\endgroup$ – gnasher729 Dec 21 '17 at 20:05
  • $\begingroup$ I was thinking of FLOPS which, afaik, are measured/calculated assuming no I/O happens. Which is purely theoretical, of course, but still. $\endgroup$ – Raphael Dec 21 '17 at 21:12
  • $\begingroup$ This answer would be fine if it had some justification. As always, we're looking for answers that explain why they're right. $\endgroup$ – David Richerby Dec 21 '17 at 22:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.